(i) Suppose that \(\mu \)  is a \(2/p\) -Carleson measure. By Proposition 1 in [1]}, we obtain that there exists a positive constant \(C\)  which is independent of \(n,~k\)  such that
\(|\mu _{n,k,\alpha }|\le C\frac{\Gamma (n+\alpha )}{n!\Gamma (\alpha )}\frac{1}{(k+1)^{2/p}}.\) 
This and Theorem 4 in [2]} imply that, for every \(n\) ,
\(\sum _{k=0}^{\infty }\left|\mu _{n, k,\alpha }\right|\left|a_{k}\right| & \le C\frac{\Gamma (n+\alpha )}{n!\Gamma (\alpha )} \sum _{k=0}^{\infty } \frac{\left|a_{k}\right|^{p}\left|a_{k}\right|^{1-p}}{(k+1)^{2/p}} \\& \le C\frac{\Gamma (n+\alpha )}{n!\Gamma (\alpha )} \sum _{k=0}^{\infty } \frac{\left|a_{k}\right|^{p}\left|k+1\right|^{\frac{(2-p)(1-p)}{p}}}{(k+1)^{2/p}}\\&\le C\frac{\Gamma (n+\alpha )}{n!\Gamma (\alpha )} \sum _{k=0}^{\infty } (k+1)^{p-3}\left|a_{k}\right|^{p}.\) 

This converges by Theorem 3 in [2]}. So the power series in (REF ) is well defined analytic function in \(\mathbb {D}\)  and
\(\sum _{k=0}^{\infty } \mu _{n, k,\alpha } a_{k}=\int _{[0,1)}\frac{\Gamma (n+\alpha )}{n!\Gamma (\alpha )}t^n f(t)d\mu (t).\) 
On the other hand, notice that for each \(z\in \mathbb {D}\) ,
\(&\sum _{n=0}^{\infty }(\int _{[0,1)}\frac{\Gamma (n+\alpha )}{n!\Gamma (\alpha )}t^n |f(t)|d\mu (t))|z|^n\\\le & \sum _{n=0}^{\infty }\left(\sum _{k=0}^{\infty } \mu _{n, k}|a_{k}|\right)|z|^n\le C \sum _{k=0}^{\infty }\left|a_{k}\right|^{p} (k+1)^{p-3}\sum _{n=0}^{\infty }\frac{\Gamma (n+\alpha )}{n!\Gamma (\alpha )}|z|^n\\\le &\frac{C}{(1-|z|)^{\alpha }}\sum _{k=0}^{\infty }\left|a_{k}\right|^{p} (k+1)^{p-3}<\infty .\) 

Therefore, we have
\(\mathcal {H}_{\mu ,\alpha }(f)(z) &=\sum _{n=0}^{\infty }\left(\int _{[0,1)}\frac{\Gamma (n+\alpha )}{n!\Gamma (\alpha )} t^{n} f(t) d \mu (t)\right) z^{n} \\&=\int _{[0,1)} \sum _{n=0}^{\infty }\frac{\Gamma (n+\alpha )}{n!\Gamma (\alpha )}(t z)^{n} f(t) d \mu (t) \\&=\int _{[0,1)} \frac{f(t)}{(1-t z)^{\alpha }} d \mu (t)= \mathcal {I}_{\mu ,\alpha }(f)(z),\) 

for every \(z\in \mathbb {D}\)  and \(f\in A^p\) .
The proof of (ii) and (iii) is analogous to the proof of (i). We omit the details.
Notice that \({B}_{\beta }\subset \cap _{0<p<\infty }A^{p}\)  for \(0<\beta \le 1\) . We can easily obtain the following corollary by Proposition  REF
Corollary 3.1 
Suppose \(\alpha >0\) , \(0<\beta \le 1\) . If \(\mu \)  is a \(t\) -Carleson measure for some \(t>0\) , then for every \(f\in {B}_{\beta }\)  and \(z\in \mathbb {D}\) ,
\( \mathcal {H}_{\mu ,\alpha }(f)(z)=\mathcal {I}_{\mu ,\alpha }(f)(z).\) 

Proposition 3.2 
Suppose \(\alpha ,\beta >0\) , if \(\mu \)  is a \(\beta \) -Carleson measure, then for every \(f\in {B}_{\beta }\)  and \(z\in \mathbb {D}\) ,
\( \mathcal {H}_{\mu ,\alpha }(f)(z)=\mathcal {I}_{\mu ,\alpha }(f)(z).\) 

To proof the Proposition REF , we still need the following lemma, which can be found in [4]}.
Lemma 3.1 
Let \(f(z)=\sum _{n=0}^{\infty }a_{n}z^{n}\in {B}_{\alpha }\)  for any \(\alpha >0\) . Then
\(\sup _{n}\sum _{k=2^{n}+1}^{2^{n+1}}\big |\frac{a_{k}}{k^{\alpha -1}}\big |^{2}<C\Vert f\Vert _{{B}_{\alpha }}^{2}.\) 

Proof of Proposition REF . Since \(\mu \)  is a \(\beta \) -Carleson measure, by Proposition 1 in [1]}, we have
\(|\mu _{n,k,\alpha }|\le C\frac{\Gamma (n+\alpha )}{n!\Gamma (\alpha )}\frac{1}{(k+1)^{\beta }}.\) 
Notice that \(\frac{\Gamma (n+\alpha )}{n!\Gamma (\alpha )}\sim n^{\alpha -1}\)  as \(n\rightarrow \infty \)  by Stirling's formula. It follows that there exist a constant \(C>0\)  such that
\(\sum _{n=0}^{\infty }\left(\sum _{k=2^{n}+1}^{2^{n+1}}\big |\frac{\mu _{n,k,\alpha }}{k^{1-\beta }}\big |^{2}\right)^{1/2}\le C\sum _{n=0}^{\infty }\left(n^{\alpha -1}\sum _{k=2^{n}+1}^{2^{n+1}}\frac{1}{k^2}\right)^{1/2}\le C\sum _{n=0}^{\infty }\frac{n^{(\alpha -1)/2}}{2^{n/2}}<\infty .\) 

This together with Lemma REF  yield that for every \(f\in {B}_{\beta }\) ,
\(\sum _{k=1}^{\infty }|\mu _{n,k,\alpha }a_k|&\le \sum _{k=1}^{\infty }\frac{\mu _{n,k,\alpha }}{k^{1-\beta }}\frac{|a_{k}|}{k^{\beta -1}}\\&\le \sum _{n=0}^{\infty }\left(\sum _{k=2^{n}+1}^{2^{n+1}}\big |\frac{\mu _{n,k,\alpha }}{k^{\beta -1}}\big |^{2}\right)^{1/2}\sup _{n}\left(\sum _{k=2^{n}+1}^{2^{n+1}}\big |\frac{a_{k}}{k^{\beta -1}}\big |^{2}\right)^{1/2}\\&<C\Vert f\Vert _{{B}_{\beta }}.\) 

From now on, arguing as in Proposition REF  (i), we can obtain the desired result.
Our next objective is to give a necessary condition such that the operator \(\mathcal {H}_{\mu ,\alpha }\)  is bounded on Bloch type spaces for general cases.
Theorem 3.1 
Suppose \(\alpha ,\beta ,\gamma >0\)  and \(\mathcal {H}_{\mu ,\alpha }\)  is a bounded operator from \({B}_{\beta }\)  into \({B}_{\gamma }\) .
(\(i\) )
If \(1<\beta <\infty \) , then the measure \(\mu \)  is an \((\alpha +\beta -\gamma -1/2)\) -Carleson measure if \(\alpha +\beta -\gamma -1/2>0\)  and \(\mu \)  is finite if \(\alpha +\beta -\gamma -1/2= 0\) .
(\(ii\) )
If \(0<\beta <1\) , then the measure \(\mu \)  is an \((\alpha -\gamma +1/2)\) -Carleson measure if \(\alpha -\gamma +1/2>0\)  and \(\mu \)  is finite if \(\alpha -\gamma +1/2= 0\) .


